The period of a simple pendulum is measured to be 4.0 seconds in a frame of reference. Which of the following is correct when the period of the pendulum is measured by an observer moving with 0.4 times the speed of the light?

Choose one answer.
a. The time period of the pendulum remains constant.
b. The time period of the pendulum decreases.
c. The time period of the pendulum increases.
d. The time period of the pendulum decreases by half.

it remains constant for that observer.

To answer this question, we need to consider the effects of time dilation due to the observer's high speed.

According to Einstein's theory of relativity, time dilation occurs when an object is in motion relative to an observer. As the observer approaches the speed of light, time appears to slow down for them compared to a stationary observer.

In this scenario, the observer is moving at 0.4 times the speed of light. We can calculate the time dilation factor using the formula:

Time dilation factor = 1 / √(1 - v^2/c^2),

where v is the velocity of the observer and c is the speed of light.

Plugging in the values, we get:

Time dilation factor = 1 / √(1 - (0.4)^2),

Time dilation factor = 1 / √(1 - 0.16),

Time dilation factor = 1 / √(0.84),

Time dilation factor = 1 / 0.916,

Time dilation factor = 1.092.

Now, to find the new period of the pendulum as measured by the moving observer, we can multiply the original period by the time dilation factor:

New period = Original period x Time dilation factor,

New period = 4.0 seconds x 1.092.

Calculating, we find:

New period ≈ 4.368 seconds.

Comparing the original period and the new period, we can see that the time period of the pendulum increases when measured by the observer moving at 0.4 times the speed of light.

Therefore, the correct answer is c. The time period of the pendulum increases.